Precalculus Examples

$x^{2} + 21 > 10 x$

Step 1

Subtract $10 x$ from both sides of the inequality.

$x^{2} + 21 - 10 x > 0$

Step 2

Convert the inequality to an equation.

$x^{2} + 21 - 10 x = 0$

Step 3

Factor $x^{2} + 21 - 10 x$ using the AC method.

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Step 3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $21$ and whose sum is $- 10$ .

$- 7, - 3$

Step 3.2

Write the factored form using these integers.

$(x - 7) (x - 3) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 7 = 0$

$x - 3 = 0$

Step 5

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 5.2

Add $7$ to both sides of the equation.

$x = 7$

Step 6

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 6.2

Add $3$ to both sides of the equation.

$x = 3$

Step 7

The final solution is all the values that make $(x - 7) (x - 3) = 0$ true.

$x = 7, 3$

Step 8

Use each root to create test intervals.

$x < 3$

$3 < x < 7$

$x > 7$

Step 9

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 9.1

Test a value on the interval $x < 3$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval $x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 9.1.2

Replace $x$ with $0$ in the original inequality.

${(0)}^{2} + 21 > 10 (0)$

Step 9.1.3

The left side $21$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.2

Test a value on the interval $3 < x < 7$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $3 < x < 7$ and see if this value makes the original inequality true.

$x = 5$

Step 9.2.2

Replace $x$ with $5$ in the original inequality.

${(5)}^{2} + 21 > 10 (5)$

Step 9.2.3

The left side $46$ is not greater than the right side $50$ , which means that the given statement is false.

False

Step 9.3

Test a value on the interval $x > 7$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

Step 9.3.2

Replace $x$ with $10$ in the original inequality.

${(10)}^{2} + 21 > 10 (10)$

Step 9.3.3

The left side $121$ is greater than the right side $100$ , which means that the given statement is always true.

True

Step 9.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < 3$ True

$3 < x < 7$ False

$x > 7$ True

$x < 3$ True

$3 < x < 7$ False

$x > 7$ True

Step 10

The solution consists of all of the true intervals.

$x < 3$ or $x > 7$

Step 11

Convert the inequality to interval notation.

$(- \infty, 3) \cup (7, \infty)$

Step 12